Vtyjkyuk

We know that if there is a curve $$Gamma=(x,y)inBbb R^2 : y=f(x), xin[a,b]$$ then
$$textlength(Gamma)=intlimits_a^bsqrt1+f'(t)^2dt$$
and I get that this is because
$$textlength(Gamma)=intlimits_Gamma1=intlimits_a^b1cdot|gamma'(t)|dt=intlimits_a^b|(gamma_1'(t),gamma_2'(t))|dt=intlimits_a^bsqrtgamma_1'(t)^2+gamma_2'(t)^2dt$$ where $gamma(t)=(gamma_1(t),gamma_2(t))=(t,f(t))$

But $sqrt1+f'(t)^2$ is another function in itself and I would like to see what it corresponds to, is there a way of comparing the two functions in a more meaningful way then just the reasoning that I wrote above?

I am looking for an animation or a drawing or simply an explanation.

The formula arises from an analysis of the curve $(t,f(t))$. Suppose you move from
a point $p_1=(t_1,f(t_1))$ to a point $p_2(t_2,f(t_2))$ on the curve, with $t_1<t_2$. If $t_1,t_2$ are sufficiently close to each other, the distance between these two points is close to the length of the piece of the curve between the points $p_1,p_2$. Now the distance is simply:
$$d(p_1,p_2)=sqrt(t_2-t_1)^2+(f(t_2)-f(t_1))^2$$
Assuming $f$ is differentiable, the mean value theorem guarantees that we can find a point $xiin(t_1,t_2)$ such that
$$f(t_2)-f(t_1)=f'(xi)(t_2-t_1)$$
so we can write
$$d(p_1,p_2)=sqrt(t_1-t_2)^2+(f^'(xi))^2(t_2-t_1)^2=(t_2-t_1)sqrt1+(f'(xi))^2$$
Now if we partition the curve into points $(t_1,f(t_1)),(t_2,f(t_2)),dots,(t_n,f(t_n))$ and add all these distances we just calculated, we obtain the sum
$$sum_i=1^n(t_i+1-t_i)sqrt(f'(xi_i))^2+1$$
which is a Riemman sum for the integral $int_a^bsqrt1+(f'(x))^2,dx$.

If $gamma(t)=(x(t),y(t))$ then $gamma'(t)=(x'(t),y'(t))$ and $gamma'(t)dt = (dx,dy)$, so the infinitesimal element of length is $sqrtdx^2+dy^2 = |gamma'(t)|,dt$ and $|gamma'(t)|=sqrtleft(fracdxdtright)^2+left(fracdydtright)^2$. If $gamma$ is the graph of a function we have $x(t)=t$ and $y(t)=f(t)$, so $|gamma'(t)|=sqrt1+f'(x)^2$. Long story short, they are exactly the same formula.

I did not feel like drawing so I'm just going to write.

Imagine the graph of a function, and on it two points $f(x)$ and $f(x+Delta x)$. Sketch lines through these points that are parallel to the axes - this will form two right triangles which are essentially the same.

On both of these triangles, one of the legs has length $Delta x$, and the other $vert f(x)-f(x+Delta x) vert.$ By Pythagoras, the hypothenuse therefore has length
$$sqrt(Delta x)^2 + (f(x)-f(x+Delta x))^2.$$
The hypothenuse is a good approximation of the length of the curve generated by $f$ between $x$ and $x+Delta x$.

Now,
$$sqrt(Delta x)^2 + (f(x)-f(x+Delta x))^2 = Delta x sqrt1 + left(fracf(x)-f(x+Delta x)Delta xright)^2,$$
which becomes
$$ sqrt1+(f')^2;dx $$
when you let $Delta x to 0$.
So you can think of this function as an infinitesimal length of the curve $f$ gives rise to.